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3 years ago

Most chemical reactions can be divided into how many main groups?

1 answer:

5 0

There is synthesis
decomposition
double displacement
single displacement
combustion
metathesis
so i guess you could say 6

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A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite

DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x 4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x 4 π

≥ . 5254 x ⁻²⁵

h / λ ≥ . 5254 x ⁻²⁵

6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ

12.56 x 10⁻⁹ ≥ λ

longest wave length = 12.56 n m

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3 years ago

Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

laiz [17]

Answer:

Explanation:

AP3X

5 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

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3 years ago

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To get you to aooreciate the benefits of atomic reaserch

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